This problem analyzes a simple pendulum's behavior under varying gravitational acceleration. The formula for the time period \( T \) of a simple pendulum is given by \( T = 2\pi \sqrt{\frac{L}{g}} \), where \( L \) represents the pendulum's length and \( g \) is the acceleration due to gravity.
On Earth, the gravitational acceleration \( g \) is calculated as \( \frac{GM_e}{R_e^2} \), with \( G \) being the gravitational constant, \( M_e \) the Earth's mass, and \( R_e \) the Earth's radius.
For a planet with a mass of \( 4M_e \) and a radius of \( 2R_e \), the gravitational acceleration \( g' \) is derived as follows:
\( g' = \frac{G \cdot 4M_e}{(2R_e)^2} = \frac{4GM_e}{4R_e^2} = \frac{GM_e}{R_e^2} = g \)
Consequently, the time period \( T' \) on this planet is:
\( T' = 2\pi \sqrt{\frac{L}{g'}} = 2\pi \sqrt{\frac{L}{g}} = T \)
This demonstrates that the time period remains identical on both Earth and the planet, substantiating assertion (A).
Now, examining reason (R), which states that the pendulum's mass is constant in both locations. Although true, the pendulum's mass does not influence its time period \( T \), as indicated by the formula. Therefore, (R) does not provide an explanation for (A).
The correct conclusion is: (A) is true but (R) is false.
Net gravitational force at the center of a square is found to be \( F_1 \) when four particles having masses \( M, 2M, 3M \) and \( 4M \) are placed at the four corners of the square as shown in figure, and it is \( F_2 \) when the positions of \( 3M \) and \( 4M \) are interchanged. The ratio \( \dfrac{F_1}{F_2} = \dfrac{\alpha}{\sqrt{5}} \). The value of \( \alpha \) is 
